\(\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [558]

Optimal result

Integrand size = 38, antiderivative size = 214 \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{3/2} d}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(11 A+5 i B) \sqrt {\cot (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(25 A+7 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{6 a^2 d} \]

[Out]

(1/4+1/4*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+
c)^(1/2)/a^(3/2)/d+1/6*(11*A+5*I*B)*cot(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^(1/2)-1/6*(25*A+7*I*B)*cot(d*x+c)^
(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^2/d+1/3*(A+I*B)*cot(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4326, 3677, 3679, 12, 3625, 211} \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {(25 A+7 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{6 a^2 d}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(11 A+5 i B) \sqrt {\cot (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Int[(Cot[c + d*x]^(3/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((1/4 + I/4)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d
*x]]*Sqrt[Tan[c + d*x]])/(a^(3/2)*d) + ((A + I*B)*Sqrt[Cot[c + d*x]])/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((1
1*A + (5*I)*B)*Sqrt[Cot[c + d*x]])/(6*a*d*Sqrt[a + I*a*Tan[c + d*x]]) - ((25*A + (7*I)*B)*Sqrt[Cot[c + d*x]]*S
qrt[a + I*a*Tan[c + d*x]])/(6*a^2*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3625

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-2*a*(b/f), Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3677

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*
f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3679

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*
(n + 1)*(c^2 + d^2))), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx \\ & = \frac {(A+i B) \sqrt {\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{2} a (7 A+i B)-2 a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2} \\ & = \frac {(A+i B) \sqrt {\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(11 A+5 i B) \sqrt {\cot (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{4} a^2 (25 A+7 i B)-\frac {1}{2} a^2 (11 i A-5 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{3 a^4} \\ & = \frac {(A+i B) \sqrt {\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(11 A+5 i B) \sqrt {\cot (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(25 A+7 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{6 a^2 d}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {3 a^3 (i A+B) \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{3 a^5} \\ & = \frac {(A+i B) \sqrt {\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(11 A+5 i B) \sqrt {\cot (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(25 A+7 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{6 a^2 d}+\frac {\left ((i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{4 a^2} \\ & = \frac {(A+i B) \sqrt {\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(11 A+5 i B) \sqrt {\cot (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(25 A+7 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{6 a^2 d}-\frac {\left (i (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{2 d} \\ & = \frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{3/2} d}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(11 A+5 i B) \sqrt {\cot (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(25 A+7 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{6 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.15 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.92 \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {i \left (3 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (i+\cot (c+d x)) \sqrt {a+i a \tan (c+d x)}+\frac {2 a (-39 i A+9 B-12 A \cot (c+d x)+(25 A+7 i B) \tan (c+d x))}{\sqrt {i a \tan (c+d x)}}\right )}{12 d \cot ^{\frac {5}{2}}(c+d x) (i a \tan (c+d x))^{3/2} (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[(Cot[c + d*x]^(3/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((I/12)*(3*Sqrt[2]*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*(I + Cot[c +
d*x])*Sqrt[a + I*a*Tan[c + d*x]] + (2*a*((-39*I)*A + 9*B - 12*A*Cot[c + d*x] + (25*A + (7*I)*B)*Tan[c + d*x])
)/Sqrt[I*a*Tan[c + d*x]]))/(d*Cot[c + d*x]^(5/2)*(I*a*Tan[c + d*x])^(3/2)*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan
[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 938 vs. \(2 (172 ) = 344\).

Time = 0.61 (sec) , antiderivative size = 939, normalized size of antiderivative = 4.39

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(939\)
default	\(\text {Expression too large to display}\)	\(939\)

[In]

int(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/24/d*(1/tan(d*x+c))^(3/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*(3*I*B*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^4+9*I*A*ln(-(-2*2^(1/2)
*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^3
-3*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c
)+I))*a*tan(d*x+c)^4-9*I*B*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+
c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^2+28*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c
)^3+9*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*
x+c)+I))*a*tan(d*x+c)^3-3*I*A*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d
*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)-256*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x
+c)^2+9*A*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I
))*2^(1/2)*a*tan(d*x+c)^2+100*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3-36*I*B*(-I*a)^
(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-3*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)+64*B*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan
(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+48*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-204*A*(-I*a)^(1/2)*t
an(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/a^2/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^3/(
-I*a)^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 465 vs. \(2 (162) = 324\).

Time = 0.26 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.17 \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left (2 \, {\left (19 \, A + 4 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (13 \, A + 7 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

[In]

integrate(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt(1/2)*a^2*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2)*sqrt(1/2)
*(I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e
^(2*I*d*x + 2*I*c) - 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^3*d^2)) + (A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*
c)/(I*A + B)) - 3*sqrt(1/2)*a^2*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2)*
sqrt(1/2)*(-I*a^2*d*e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*
c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^3*d^2)) + (A - I*B)*a*e^(I*d*x + I*c))*e^(-
I*d*x - I*c)/(I*A + B)) + sqrt(2)*(2*(19*A + 4*I*B)*e^(4*I*d*x + 4*I*c) - (13*A + 7*I*B)*e^(2*I*d*x + 2*I*c) -
A - I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-3
*I*d*x - 3*I*c)/(a^2*d)

Sympy [F]

\[ \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(cot(d*x+c)**(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral((A + B*tan(c + d*x))*cot(c + d*x)**(3/2)/(I*a*(tan(c + d*x) - I))**(3/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

\[ \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{\frac {3}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*cot(d*x + c)^(3/2)/(I*a*tan(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

[In]

int((cot(c + d*x)^(3/2)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int((cot(c + d*x)^(3/2)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^(3/2), x)